Using factor theorem, factorize the following:

2x³ + 19x² + 38x + 21

Let, f(x) = 2x³ + 19x² + 38x + 21.

Substituting, x = −1 in f(x), we get :

f(-1) = 2(-1)³ + 19(-1)² + 38(-1) + 21

= 2(-1) + 19(1) - 38 + 21

= -2 + 19 - 38 + 21

= 0.

Since, f(−1) = 0, (x + 1) is a factor of f(x).

Dividing f(x) by (x + 1), we get :

$\begin{array}{l} \phantom{x - ]3)}{2x^2 + 17x + 21} \ x + 1\overline{\smash{\big)}2x^3 + 19x^2 + 38x + 21} \ \phantom{x - 2}\underline{\underset{-}{}2x^3 \underset{-}{+}2x^2} \ \phantom{{x - 2}x^,,,3-}17x^2 + 38x \ \phantom{{x -l2}fx^3]\space}\underline{\underset{-}{+}17x^2 \underset{-}{+}17x} \ \phantom{{x - ll2]euo[ki]}x^3okk\space}{21x + 21} \ \phantom{{x - 2}x^3o;lklk]lmk\space}\underline{\underset{-}{+}21x\underset{-}{+} 21} \ \phantom{{x - 2}{x^,jo-k2x^2k\space}{-9x}}\times \end{array}$

∴ 2x³ + 19x² + 38x + 21 = (x + 1)(2x² + 17x + 21)

= (x + 1)(2x² + 14x + 3x + 21)

= (x + 1)[2x(x + 7) + 3(x + 7)]

= (x + 1)(x + 7)(2x + 3)

Hence, 2x³ + 19x² + 38x + 21 = (x + 1)(x + 7)(2x + 3).